Problem: What is the extraneous solution to these equations? $\dfrac{x^2 - 13}{x + 4} = \dfrac{-9x - 33}{x + 4}$
Solution: Multiply both sides by $x + 4$ $ \dfrac{x^2 - 13}{x + 4} (x + 4) = \dfrac{-9x - 33}{x + 4} (x + 4)$ $ x^2 - 13 = -9x - 33$ Subtract $-9x - 33$ from both sides: $ x^2 - 13 - (-9x - 33) = -9x - 33 - (-9x - 33)$ $ x^2 - 13 + 9x + 33 = 0$ $ x^2 + 20 + 9x = 0$ Factor the expression: $ (x + 5)(x + 4) = 0$ Therefore $x = -5$ or $x = -4$ At $x = -4$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -4$, it is an extraneous solution.